# quadratic equations

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• February 22nd 2008, 07:28 PM
andrew2322
quadratic equations
hi all, i need some help on finding quadratic equations from points on graphs, could someone please please explain how i would determine which general equation to use to solve.

1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.

a.) someone please explain how to pick a formula to use

2.) a parabola passes through points (-3,9) find the rule dictating the parabola.

a.) someone please explain how to pick a formula to use

also at the start of the exercise there is an example like this: a parabola passes through points (0,3) and (-3,1) it says this equation is in the from y =ax^2 + c... What the hell? i nvr learnt any ax^2 +c

pls help i have a test this coming thursday
• February 22nd 2008, 07:45 PM
mr fantastic
Quote:

Originally Posted by andrew2322
hi all, i need some help on finding quadratic equations from points on graphs, could someone please please explain how i would determine which general equation to use to solve.

1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.

Mr F says: Use the model $y = A(x - B)(x - C)$. B = 1, C = 3 (why?). To get A substitute the given y-intercept, that is, the point (0, 3).

a.) someone please explain how to pick a formula to use

2.) a parabola passes through points (-3,9) find the rule dictating the parabola. Mr F says: Insufficient data to get a unique rule.

a.) someone please explain how to pick a formula to use

also at the start of the exercise there is an example like this: a parabola passes through points (0,3) and (-3,1) it says this equation is in the from y =ax^2 + c... What the hell? i nvr learnt any ax^2 +c

Mr F says: They're giving you the heads-up that b = 0 in the standard form! Run with the ball!! Substitute the two given points to get two equations in a and c. Solve these two equations simultaneously.

pls help i have a test this coming thursday

..
• February 22nd 2008, 07:50 PM
andrew2322
sorry
huh? im like so lost... how can y = A(x-c)(x-b) work?gah pls go over it again with the actual answers pls? nd how to select which formula to use pls
• February 22nd 2008, 07:58 PM
mr fantastic
Quote:

Originally Posted by andrew2322
huh? im like so lost... how can y = A(x-c)(x-b) work?gah pls go over it again with the actual answers pls? nd how to select which formula to use pls

Consider ....... y = A(x - B)(x - C) has x-intercepts found from the equation

A(x - B)(x - C) = 0

=> x - B = 0 or x - C = 0

=> x = B or x = C.

So if you know the x-intercepts, you use the model y = A(x - B)(x - C) because knowing the x-intercepts means you know the value of B and C .....

So for the question:

Quote:

Originally Posted by andrew2322
1.) A parabola has a y intercept of 3 and x intercepts of 1 and 3, find the rule dictating the parabola.

you would use this model and you know that B = 1 and C = 3.

So y = A(x - 1)(x - 3).

Now substitute the y-inetercept, that is, the point (0, 3), that is, x = 0 and y = 3:

3 = A(0 - 1)(0 - 3) = A(-1)(-3) = 3A => A = 1.

So y = (x - 1)(x - 3).
• February 22nd 2008, 08:01 PM
andrew2322
woah
wow that just hit me in the face, how cud i have not seen the obvious... wow ur really smart. and one last thing, cud u pls explain the values B and C and how they are different from X? and also, there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.
• February 22nd 2008, 08:04 PM
mr fantastic
Quote:

Originally Posted by andrew2322
wow that just hit me in the face, how cud i have not seen the obvious... wow ur really smart. and one last thing, cud u pls explain the values B and C and how they are different from X?

They are the x-intercepts. In other words, the value of x where the parabola cuts the x-axis. In other words, when y = 0, x = B or x = C.

Try working backwards from the answer y = (x - 1)(x - 3)to see how it all fits together.
• February 22nd 2008, 08:04 PM
andrew2322
sorry
there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.
• February 22nd 2008, 08:07 PM
mr fantastic
Quote:

Originally Posted by andrew2322
there is another example as y = ax^2? how did that come about? the question is the parabola passes through points (2,5) find the equation that dictates the graph.

They give you the model - on a platter. Don't question your good luck, just use it!!!!

Substitute the given point, that is x = 2 and y = 5:

5 = a (2)^2 = 4a => a = 5/4.

Therefore y = (5/4) x^2.
• February 22nd 2008, 08:09 PM
andrew2322
hey
could you please tell me how, if i didnt have good luck, know what equation to use?
• February 22nd 2008, 08:13 PM
mr fantastic
Quote:

Originally Posted by andrew2322
could you please tell me how, if i didnt have good luck, know what equation to use?

You wouldn't. Without being given the model, a single point is insufficient data to get a unique equation. That's why you were given the equation (so it wasn't really good luck .... it was actually absolutely essential).
• February 22nd 2008, 08:18 PM
andrew2322
one last
ok just one FINAL question

Y intercept: (0,5)
X intercept: (4,0)

Find the equation, please give me step by step working out with the final answer, thanks
• February 22nd 2008, 08:59 PM
mr fantastic
Quote:

Originally Posted by andrew2322
ok just one FINAL question

Y intercept: (0,5)
X intercept: (4,0)

Find the equation, please give me step by step working out with the final answer, thanks

Insufficient data to find a unique equation. Is the x-intercept also a turning point?